Questions
MAT136H5 S 2025 - All Sections 6.1 preparation check
Multiple dropdown selections
In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 8n n! xn Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| 8n+1xn+1 (n+1)! 8nxn n! | [ Select ] 1 / 4 0 1 infinity 8 1 / 8 b) What does the Ratio Test and your answer in (a) tell you about the series? [ Select ] The series converges only when -1 < x < 1 The series converges only when -8 < x < 8 The series converges for all values of x The series only converges when x = 0 The series does not converge for any values of x c) What is the radius of convergence for the series? R = infinity d) What is the interval of convergence for the series? [ Select ] (-infinity, infinity) [-1, 1] (-8, 8) [-8, 8] x = 0 (-1, 1) Note: The interval of convergence is the set of values of x for which the series converges. e) Consider your answer to (d). Does the series converge when x=3 ? [ Select ] Yes There is not enough information No
View Explanation
Verified Answer
Please login to view
Step-by-Step Analysis
As we work through the given power series, a clear path is to recognize its form. The general term is a_n = (8^n / n!) x^n, so the series is sum_{n=0}^∞ (8^n x^n)/n! = sum_{n=0}^∞ ( (8x)^n / n!).
Option by option reasoning:
Option a) The choice '0' for lim_{n→∞} |a_{n+1}/a_n|' obviously targets the ratio test limit. Compute the ratio: |a_{n+1}/a_n| = | (8^{n+1} x^{n+1}/(n+1)!) / (8^n x^n / n!) | = | (8x)/(n+1) |.
As n → ∞, this tends to 0 for every fixed x, since the denominator grows without bound while the numerator stays constant. Hence the ......Login to view full explanationLog in for full answers
We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!
Similar Questions
Question text 5Marks a) Find the radii of convergence of the following series:[math: ∑n=1∞n2+2n3n2+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3n^2+2}x^n radius = Answer 1[input][math: ∑n=1∞n2+2n3n+2xn]\displaystyle \sum_{n=1}^{\infty} \dfrac{n^2+2n}{3^n+2}x^n radius = Answer 2[input]b) Suppose a power series [math: ∑n=1∞anxn]\displaystyle \sum_{n=1}^{\infty} a_n x^n is convergent for [math: x=−3]x= -3 and divergent for [math: x=5]x = 5.For [math: x=−1]x = -1, the power series Answer 3[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=−6]x = -6, the power series Answer 4[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]For [math: x=3]x = 3, the power series Answer 5[select: , is convergent., is divergent., can be convergent or divergent depending on the coefficients.]Notes Report question issue Question 8 Notes
Which of the following statements is not correct?
Which of the following statements is NOT true?
In this question we work through the steps for solving the following problem: Find the interval of convergence and the radius of convergence of the power series ∞ ∑ n=0 (x−4)n 3n2 Solution outline: To find interval and radius of convergence, we usually start by using the Ratio Test: a) Evaluate the limit lim n→∞| an+1 an |=lim n→∞| (x−4)n+1 3(n+1)2 (x−4)n 3n2 | |x-4| b) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|<1 ? [ Select ] Nothing (because the ratio test inconclusive) The power series converges The power series diverges c) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|>1 ? [ Select ] The power series converges Nothing (because the ratio test inconclusive) The power series diverges d) What does the Ratio Test and your answer in (a) tell you about convergence when |x−4|=1 ? [ Select ] The power series diverges The power series converges Nothing (because the ratio test inconclusive) e) For what values of x is |x−4|<1 ? (Solve the inequality) [ Select ] x < 5 -4 < x < 5 -4 < x < 4 3 < x < 5 2 < x < 6 f) When x=5 the power series in the question becomes ∞ ∑ n=0 1 3n2 . Does this series converge or diverge? [ Select ] Diverges Converges g) When x=3 the power series in the question becomes ∞ ∑ n=0 (−1)n 3n2 . Does this series converge or diverge? [ Select ] Diverges Converges h) Summarizing the results we have found, for what values of x does the power series in the question converge? [ Select ] Converges for x in the interval (-4,4) Converges for x in the interval [3,5] Converges for x in the interval (-4,5] Converges for x in the interval [3,5) Converges for x in the interval (3,5] Converges for x in the interval [-4,4] Note: This is the interval of convergence.
More Practical Tools for Students Powered by AI Study Helper
Making Your Study Simpler
Join us and instantly unlock extensive past papers & exclusive solutions to get a head start on your studies!